Heat stress in dairy cows. Reaserchers in the IFAS DairyResearch Unit and the De

Question

Heat stress in dairy cows. Reaserchers in the IFAS DairyResearch Unit and the Department of Agricultural Engineering at theUniversity of Florida have developed design criteria for theconstruction of shade structures that they believe will helpalleviate heat stress for dariy cows. In one experiment, 31Holstein cows in the last trimester of pregnancy were divided intotwo groups. Sixteen cows were given access to a shade structure andthe remaining 15 cows were denied shade. Researchers recorded the100 day milk yield (in pounds) of each cow after calving. The meanmilk yields of the two groups are shown in the accompanying table.Is there sufficient evidence to indicate a difference between themean milk yields of cows given access to shade and cows deniedshade? Use = .10. (Assume the standard deviations of milkyields are equal to 40 pounds for both groups.)                         Shade            NoShade SampleSize         16                  15 Mean                  367.4            330.8 Heat stress in dairy cows. Reaserchers in the IFAS DairyResearch Unit and the Department of Agricultural Engineering at theUniversity of Florida have developed design criteria for theconstruction of shade structures that they believe will helpalleviate heat stress for dariy cows. In one experiment, 31Holstein cows in the last trimester of pregnancy were divided intotwo groups. Sixteen cows were given access to a shade structure andthe remaining 15 cows were denied shade. Researchers recorded the100 day milk yield (in pounds) of each cow after calving. The meanmilk yields of the two groups are shown in the accompanying table.Is there sufficient evidence to indicate a difference between themean milk yields of cows given access to shade and cows deniedshade? Use = .10. (Assume the standard deviations of milkyields are equal to 40 pounds for both groups.)                         Shade            NoShade SampleSize         16                  15 Mean                  367.4            330.8

Explanation / Answer

Given s=40 =0.1, Z(0.05)=1.645 (check normal table) The test hypothesis is Ho:1=2 Ha:1 not equal 2 The test statistic is Z=(xbar1-xbar2)/(s*(1/n1 + 1/n2)) =(367.4 - 330.8)/(40*sqrt(1/16 +1/15)) =2.55 Since Z=2.55>1.645, we reject Ho.

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